$\begingroup$I'm very new to math and proofs -- so I apologize if my math skills and vocabulary offends you.
I have a question that states: Prove that $\pi$ is a fundamental period of the tangent function.
I need to know if the proof I wrote is adequate, here is my proof:
$a$ is a Real Number. $0 < a < \pi$
$tan(s+a) = tan(s)$
The only case in which $tan(s) = 0$ is when:
$k$ is an Integer. $tan(\pi * k) = 0 $
thus, $tan(a) != 0$
Because $tan$ is the ratio of an angle $\theta$ of $sin/cos$ and because $sin$ and $cos$ are ratios of sides of a triangle in reference to the unit circle and since the unit circle has a radius of $1$ and a circumference of $2\pi$ radians, in order for periodicity to occur, the period of tan must be $(\pi * k)$, where the arc length can never be a multiple of $\pi/2$
Thanks so much for any help!
$\endgroup$2 Answers
$\begingroup$Solve $$\tan(t)=\tan(t+T).$$
You have
$$\tan(t+T)=\frac{\tan(t)+\tan(T)}{1-\tan(t)\tan(T)}=\tan(t),$$ and for $\tan(t)\tan(T)\ne1$,
$$\tan(T)=-\tan^2(t)\tan(T).$$
This is possible only when $\tan(T)=0$, for which the smallest positive solution is
$$T=\pi.$$
$\endgroup$2$\begingroup$It's certainly true that $\tan(x +\pi) = \tan x$, so $\pi$ is at least a multiple of the period. Can you show that no smaller number will do? Hint: What is the slope of $\tan$ between $\pm \pi/2$?
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